Я пытаюсь запустить приведенный ниже код, чтобы смоделировать набор P-значений с использованием обобщенной линейной модели
Однако я продолжаю получать ошибку : процедура Лапака dgesv: система точно единственное число: U [6,6] = 0
Вот код, который я пытаюсь запустить:
#which_p_value = "x1"
which_p_value = "groupcategory"
#which_p_value = "x1:groupcategory"
run_anova = FALSE
simulate_mixed_effect = TRUE
mixed_effect_sd = 3.094069
mixed_effect_sd_slope = 3.098661
library(tidyverse)
n_people <- c(2,5,10,15,20)
coef1 <- 1.61
coef2 <- -0.01
#coef3 <- 5
#coef4 <- 0
g1 = 0
g2 = 1
g3 = 2
distances <- c(60,90,135,202.5,303.75,455.625)/100
n_trials <- 35
oneto1000 <- 25
n_track_lengths <- length(distances)
groupcategory = c(rep(g1, n_track_lengths), rep(g2, n_track_lengths),rep(g3,n_track_lengths))
z = c(n_people)
emptydataframeforpowerplots = NULL
coef3s <- c(-5, -4, -3, -2,-1, 0, 1, 2, 3, 4, 5)
coef4s <- c(-1, -0.8, -0.6, -0.4, -0.2, 0, 0.2, 0.4, 0.6, 0.8, 1)
Datarray <- array(dim=c(length(coef3s), length(coef4s),length(n_people)))
coef3_counter =1
for (coef3 in coef3s) {
coef4_counter =1
for (coef4 in coef4s) {
z1_g2 <- coef1 + coef2*distances + coef3*g2 + coef4*g2*distances
z1_g3 <- coef1 + coef2*distances + coef3*g3 + coef4*g3*distances
d = NULL
pr1 = 1/(1+exp(-z1_g2))
pr2 = 1/(1+exp(-z1_g3))
counter=1
for (i in n_people) {
for (j in 1:oneto1000){
df <- c()
for (k in 1:i){
# random effect from drawing a random intercept with sd = x
if (simulate_mixed_effect){
coef1_r = rnorm(1, mean=coef1, sd=mixed_effect_sd)
coef2_r = rnorm(1, mean=coef1, sd=mixed_effect_sd_slope)
} else {
coef1_r = coef1
coef2_r = coef2
}
z_g1 <- coef1_r + coef2*distances + coef3*g1 + coef4*g1*distances
pr = 1/(1+exp(-z_g1))
z1_g2 <- coef1_r + coef2*distances + coef3*g2 + coef4*g2*distances
pr1 = 1/(1+exp(-z1_g2))
if (run_anova) {
df <- rbind(df, data.frame(x1 = c(rep(distances, 3)),
y = c(rbinom(n_track_lengths,n_trials,pr), rbinom(n_track_lengths,n_trials,pr1),rbinom(n_track_lengths,n_trials,pr2)),
groupcategory = groupcategory, id = c(rep(k,18))))
} else { # this is for glmer data organisation
for (m in 1:n_trials) {
df <- rbind(df, data.frame(x1 = c(rep(distances, 3)),
y = c(rbinom(n_track_lengths,1,pr),rbinom(n_track_lengths,1,pr1),rbinom(n_track_lengths,1,pr2)),groupcategory = groupcategory,id = c(rep(k,18))))
}
}
}
if (run_anova) {
#df_aov <- aov(y~x1*groupcategory+Error(id/(x1*groupcategory)),data=df)
#df_aov_sum <- summary(df_aov)
#pvalue <- df_aov_sum[[5]][[1]][which_p_value,"Pr(>F)"]
df_aov <- aov(y~x1*groupcategory+Error(id),data=df)
df_aov_sum <- summary(df_aov)
pvalue <- df_aov_sum[[2]][[1]][which_p_value, "Pr(>F)"]
} else { # glmer
mod_group_glmer <- glmer(y ~ x1 + groupcategory + (1+x1|id), data = df, family = "binomial")
sum <- summary(mod_group_glmer)
pvalue <- sum$coefficients[which_p_value, "Pr(>|z|)"]
}
d = rbind(d,data.frame(pvalue))
}
count <- plyr::ldply(d,function(c) sum(c<=0.05))
Datarray[coef3_counter,coef4_counter,counter] <- count$V1/oneto1000
counter = counter +1
d = NULL
}
coef4_counter = coef4_counter + 1
}
coef3_counter = coef3_counter + 1
}
Ниже приведен скрипт отладчика:
Lapack routine dgesv: system is exactly singular: U[6,6] = 0
8. stopifnot(length(value <- as.numeric(value)) == 1L)
7. nM$newf(fn(nM$xeval()))
6. (function (fn, par, lower = rep.int(-Inf, n), upper = rep.int(Inf, n), control = list()) { n <- length(par) ...
5. do.call(optfun, arglist)
4. withCallingHandlers(do.call(optfun, arglist), warning = function(w) { curWarnings <<- append(curWarnings, list(w$message)) })
3. optwrap(optimizer, devfun, start, rho$lower, control = control, adj = adj, verbose = verbose, ...)
2. optimizeGlmer(devfun, optimizer = control$optimizer[[2]], restart_edge = control$restart_edge, boundary.tol = control$boundary.tol, control = control$optCtrl, start = start, nAGQ = nAGQ, verbose = verbose, stage = 2, calc.derivs = control$calc.derivs, use.last.params = control$use.last.params)
1. glmer(y ~ x1 + groupcategory + (1 + x1 id), data = df, family = "binomial")
Кто-нибудь сможет протянуть руку помощи относительно того, как я могу действовать отсюда?